6
APPLICATIONS OF INTEGRATION
APPLICATIONS OF INTEGRATION
6.4
Work
In this section, we will learn about:
Applying integration to calculate the amount of
work done in performing a certain physical task.
WORK
The term ‘work’ is used in everyday
language to mean the total amount of
effort required to perform a task.
WORK
In physics, it has a technical meaning that
depends on the idea of a ‘force.’
 Intuitively, you can think of a force as describing a push
or pull on an object.
 Some examples are: the horizontal push of a book
across a table or the downward pull of the earth’s
gravity on a ball.
FORCE
Defn. 1 / Eqn. 1
In general, if an object moves along a straight
line with position function s(t), then:
 The force F on the object (in the same direction)
is defined by Newton’s Second Law of Motion as
the product of its mass m and its acceleration.
d 2s
F m 2
dt
FORCE
In the SI metric system, the mass is
measured in kilograms (kg), the displacement
in meters (m), the time in seconds (s), and
the force in newtons (N = kg•m/s2).
 Thus, a force of 1 N acting on a mass of 1 kg
produces an acceleration of 1 m/s2.
FORCE
In the US Customary system,
the fundamental unit is chosen to be
the unit of force, which is the pound.
WORK
Defn. 2 / Eqn. 2
In the case of constant acceleration:
 The force F is also constant and the work done
is defined to be the product of the force F and
the distance that the object moves:
W = Fd
(work = force x distance)
WORK
If F is measured in newtons and d in meters,
then the unit for W is a newton-meter called
joule (J).
If F is measured in pounds and d in feet, then
the unit for W is a foot-pound (ft-lb), which
is about 1.36 J.
WORK
Example 1
a. How much work is done in lifting a 1.2-kg
book off the floor to put it on a desk that
is 0.7 m high?
 Use the fact that the acceleration due to gravity
is g = 9.8 m/s2.
b. How much work is done in lifting a 20-lb
weight 6 ft off the ground?
WORK
Example 1 a
The force exerted is equal and opposite
to that exerted by gravity.
 Therefore, Equation 1 gives:
F = mg = (1.2)(9.8) = 11.76 N
 Thus, Equation 2 gives the work done as:
W = Fd = (11.76)(0.7) ≈ 8.2 J
WORK
Example 1 b
Here, the force is given as F = 20 lb.
 Therefore, the work done is:
W = Fd = 20 • 6 = 120 ft-lb
 Notice that in (b), unlike (a), we did not have to
multiply by g as we were given the ‘weight’ (which
is a force) and not the mass of the object.
WORK
Equation 2 defines work as long as
the force is constant.
However, what happens if the force is
variable?
WORK
Let’s suppose that the object moves
along the x-axis in the positive direction,
from x = a to x = b.
 At each point x between a and b, a force f(x) acts
on the object—where f is a continuous function.
WORK
We divide the interval [a, b] into
n subintervals with endpoints x0, x1, . . . , xn
and equal width Δx.
 We choose a sample point xi* in the i th subinterval
[xi -1, xi ].
 Then, the force at that point is f(xi*).
WORK
If n is large, then Δx is small, and since f is
continuous, the values of f don’t change very
much over the interval [xi -1, xi ].
 In other words, f is almost constant on the interval.
 So, the work Wi that is done in moving the particle
from xi -1 to xi is approximately given by Equation 2:
Wi ≈ f(xi*) ∆x
WORK
Equation 3
Thus, we can approximate
the total work by:
n
W   f ( xi *)x
i 1
WORK
It seems that this approximation
becomes better as we make n larger.
n
W   f ( xi *)x
i 1
WORK
Definition 4
So, we define the work done in moving
the object from a to b as the limit of the
quantity as n → ∞.
 As the right side of Equation 3 is a Riemann sum,
we recognize its limit as being a definite integral.
n
 Thus,
b
W  lim  f ( xi *)x   f ( x)dx
n 
i 1
a
WORK
Example 2
When a particle is located a distance x feet
from the origin, a force of x2 + 2x pounds
acts on it.
How much work is done in moving it from
x = 1 to x = 3?
WORK
Example 2
3
x
50
2
W   ( x  2 x)dx   x  
3
1 3
1
3
3
2
2
 The work done is 16 ft-lb.
3
WORK
In the next example, we use
a law from physics: Hooke’s Law.
HOOKE’S LAW
The force required to maintain a spring
stretched x units beyond its natural length is
proportional to x
f(x) = kx
where k is a positive constant (called the
spring constant).
HOOKE’S LAW
The law holds provided that x
is not too large.
WORK
Example 3
A force of 40 N is required to hold a spring
that has been stretched from its natural length
of 10 cm to a length of 15 cm.
How much work is done in stretching the
spring from 15 cm to 18 cm?
WORK
Example 3
According to Hooke’s Law, the force required
to hold the spring stretched x meters beyond
its natural length is f(x) = kx.
 When the spring is stretched from 10 cm to 15 cm,
the amount stretched is 5 cm = 0.05 m.
 This means that f(0.05) = 40, so 0.05k = 40.
40
 Therefore,
k
0.05
 800
WORK
Example 3
Thus, f(x) = 800x and the work done in
stretching the spring from 15 cm to 18 cm is:
0.08
x 
W   800 xdx  800 
0.05
2  0.05
0.08
2
2
2

 400  0.08    0.05    1.56 J


WORK
Example 4
A 200-lb cable is 100 ft long and hangs
vertically from the top of a tall building.
How much work is required to lift the cable
to the top of the building?
 Here, we don’t have a formula for the force function.
 However, we can use an argument similar to the one
that led to Definition 4.
WORK
Example 4
Let’s place the origin at the top of the
building and the x-axis pointing downward.
 We divide the cable into
small parts with length Δx.
WORK
Example 4
If xi* is a point in the i th such interval, then all
points in the interval are lifted by roughly the
same amount, namely xi*.
 The cable weighs 2 lb/foot.
 So, the weight of the i th part
is 2Δx.
WORK
Example 4
Thus, the work done on the i th part,
in foot-pounds, is:
(2x)
force
xi *  2 xi * x
distance
WORK
Example 4
We get the total work done by adding all
these approximations and letting the number
of parts become large (so ∆x→0):
n
W  lim  2 xi * x
n 

100
i 1
2 100
2 xdx  x 
0
0
 10, 000 ft-lb
WORK
Example 5
A tank has the shape of an inverted circular
cone with height 10 m and base radius 4 m.
It is filled with water to a height of 8 m.
 Find the work required to
empty the tank by pumping
all the water to the top of
the tank.
 The density of water is
1,000 kg/m3.
WORK
Example 5
Let’s measure depths from the top of the tank
by introducing a vertical coordinate line.
 The water extends from a depth
of 2 m to a depth of 10 m.
 So, we divide the interval
[2, 10] into n subintervals
with endpoints x0, x1, . . . , xn
and choose xi* in the i th
subinterval.
WORK
Example 5
This divides the water into n layers.
 The i th layer is approximated by a circular cylinder
with radius ri and height Δx.
WORK
Example 5
We can compute ri from similar triangles:
ri
4

ri  52 10  xi *
10  xi * 10
WORK
Example 5
Thus, an approximation to the volume
of the i th layer of water is:
4
2
Vi   ri x 
10  xi * x
25
2
 So, its mass is: mi = density x volume
4
2
 1000 
10  xi * x
25
 160 10  xi * x
2
WORK
Example 5
The force required to raise this layer must
overcome the force of gravity and so:
Fi  mi g
 (9.8)160 (10  xi *) x
2
 1570 (10  xi *) x
2
WORK
Example 5
Each particle in the layer must travel
a distance of approximately xi*.
 The work Wi done to raise this layer to the top
is approximately the product of the force Fi and
the distance xi*:
Wi  Fi xi *
 1570 xi *(10  xi *) x
2
WORK
Example 5
To find the total work done in emptying
the entire tank, we add the contributions
of each of the layers and then take the limit
as n →∞.
WORK
Example 5
n
W  lim 1570 xi *(10  xi *) 2 x
n 
i 1
10
  1570 xi *(10  x) dx
2
2
 1570 
10
2
2
100
x

20
x

x

 dx
10

20 x
x 
2
 1570 50 x 
 
3
4 2

3
 1570 
2048
3
  3.4 10
4
6
J